smc

Since $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1$, we can write the terms of each sequence as $$\begin{array}{rl}& \left(a_n\right)\Rightarrow \{1,x+1,2x+1,3x+1,...\}\\ & \left(b_n\right)\Rightarrow \{1,y+1,2y+1,3y+1,...\}\end{array}$$ where $x$ and $y$ ($x\le y$) are the common differences of each, respectively. Since $$\begin{array}{rl}{a}_n& =(n-1)x+1\\ b_n& =(n-1)y+1\end{array}$$ it is easy to see that $a_n\equiv b_n\equiv 1\mathrm{mod}(n-1)$. Hence, we have to find the largest $n$ such that $\frac{a_n-1}{n-1}$ and $\frac{b_n-1}{n-1}$ are both integers; equivalently, we want to maximize $\gcd (a_n-1,b_n-1)$. The prime factorization of $2010$ is $2\cdot 3\cdot 5\cdot 67$. We list out all the possible pairs that have a product of $2010$, noting that these are the possible values of $(a_n,b_n)$ and we need $a_n\le b_n$: $$(2,1005),(3,670),(5,402),(6,335),(10,201),(15,134),(30,67)$$ and soon find that the largest $n-1$ value is $7$ for the pair $(15,134)$, and so the largest $n$ value is $\boxed{8}$.